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- Thread starter nomadreid
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##X(x_1,x_2)=\langle x_1|\hat{x} x_2 \rangle=x_2 \delta(x_1-x_2).##

That's not a matrix but a distribution. The action of the operator on a vector in such a case is not given by a usual matrix-vector product but by a convolution integral. Of course, in position representation that's very simple

##\langle x|\hat{x} \psi \rangle=\int_{\mathbb{R}} \mathrm{d} x' X(x,x') \langle x'|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x' X(x,x') \psi(x')=x \psi(x)##

as it must be due to the direct evaluation

##\langle x|\hat{x} \psi \rangle = \langle \hat{x} x|\psi \rangle=x \langle x |\psi \rangle=x \psi(x).##

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Thanks very much, vanhees71. That clears up my question completely. :)

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